計算datediff時的差異MySQL

在案例中以及在where中使用datediff時，我會得到不同的結果。這是表的用戶，省略無關列，並與所有的用戶：計算datediff時的差異MySQL

<table><tbody><tr><th>UserID</th><th>UserTypeID</th><th>UserRegistrationDate</th><th>UserBirthDate</th><th>UserJobID</th><th>UserLocationID</th></tr><tr><td>1</td><td>1</td><td> </td><td> </td><td> </td><td> </td></tr><tr><td>2</td><td>1</td><td>3/12/2015</td><td>30/1/1982</td><td>2</td><td>3</td></tr><tr><td>3</td><td>1</td><td>2/12/2015</td><td>19/9/1987</td><td>2</td><td>4</td></tr><tr><td>12</td><td>1</td><td>2/12/2015</td><td>11/9/1992</td><td>2</td><td>5</td></tr><tr><td>13</td><td>1</td><td>13/12/2015</td><td>15/5/1995</td><td>2</td><td>2</td></tr><tr><td>40</td><td>1</td><td>10/1/2016</td><td>22/7/1975</td><td>2</td><td>2</td></tr><tr><td>47</td><td>1</td><td>16/1/2016</td><td>29/2/2000</td><td>1</td><td>6</td></tr><tr><td>50</td><td> </td><td> </td><td>1/12/1956</td><td>2</td><td>2</td></tr><tr><td>52</td><td> </td><td> </td><td>12/3/1998</td><td>2</td><td>6</td></tr><tr><td>54</td><td>1</td><td>20/1/2016</td><td>11/4/1990</td><td>2</td><td>2</td></tr><tr><td>55</td><td>1</td><td>25/1/2016</td><td> </td><td> </td><td> </td></tr><tr><td>57</td><td> </td><td> </td><td>1/1/1990</td><td> </td><td> </td></tr><tr><td>58</td><td>1</td><td> </td><td> </td><td> </td><td> </td></tr><tr><td>59</td><td>1</td><td>28/1/2016</td><td> </td><td> </td><td> </td></tr><tr><td>60</td><td>1</td><td>13/1/2016</td><td>1/1/1998</td><td>1</td><td>10</td></tr><tr><td>61</td><td>1</td><td>1/12/2015</td><td>22/6/1989</td><td>1</td><td>12</td></tr><tr><td>66</td><td>1</td><td>1/2/2016</td><td> </td><td> </td><td> </td></tr><tr><td>67</td><td>1</td><td>1/2/2016</td><td> </td><td> </td><td> </td></tr><tr><td>68</td><td>1</td><td>1/2/2016</td><td> </td><td> </td><td> </td></tr></tbody></table>

運行此：

SELECT 
    COUNT(users.UserID), 
    CASE WHEN users.UserBirthDate IS NULL 
    THEN 'Sin Registro' 
    WHEN datediff(now(), users.UserBirthDate)/365.25 > 50 
    THEN '> 51' 
    WHEN datediff(now(), users.UserBirthDate)/365.25 > 40 AND datediff(now(), users.UserBirthDate)/365.25 < 51 
    THEN '41 - 50' 
    WHEN datediff(now(), users.UserBirthDate)/365.25 > 30 AND datediff(now(), users.UserBirthDate)/365.25 < 41 
    THEN '31 - 40' 
    WHEN datediff(now(), users.UserBirthDate)/365.25 > 26 AND datediff(now(), users.UserBirthDate)/365.25 < 31 
    THEN '27 - 30' 
    WHEN datediff(now(), users.UserBirthDate)/365.25 > 20 AND datediff(now(), users.UserBirthDate)/365.25 < 27 
    THEN '21 - 26' 
    WHEN datediff(now(), users.UserBirthDate)/365.25 <= 20 
    THEN '< 21' END AS AgeGroup 
FROM users 
GROUP BY AgeGroup

我得到這樣的結果：

<table><tbody><tr><th>COUNT(users.userid)</th><th>AgeGroup</th></tr><tr><td>3</td><td>21 - 26</td></tr><tr><td>3</td><td>27 - 30</td></tr><tr><td>1</td><td>31 - 40</td></tr><tr><td>1</td><td>41 - 50</td></tr><tr><td>3</td><td>&lt; 21</td></tr><tr><td>1</td><td>&gt; 51</td></tr><tr><td>7</td><td>Sin Registro</td></tr></tbody></table>

但是當我運行：

SELECT COUNT(users.UserID) 
FROM users 
WHERE datediff(now(), users.UserBirthDate)/365.25 > 20 AND datediff(now(), users.UserBirthDate)/365.25 < 27;

我得到這個：

<table><tbody><tr><th>COUNT(users.userid)</th></tr><tr><td>5</td></tr></tbody></table>

UPDATE

SQLFiddle with complete tables and data.

來源

2016-02-07 RominaV

有趣。哪一個提供正確的結果？你可以使用CREATE TABLE語句和一些示例數據來設置SQL小提琴嗎？ –

只是出於好奇，如果你添加「users.UserBirthDate IS NOT NULL」到破碎的查詢會發生什麼？ – barrycarter

@AdamCopley我添加了SQLFiddle。在你的問題上「哪個是正確的？」我再次檢查，我（認爲）發現問題與查詢。謝謝！！ – RominaV

手動檢查結果後，我意識到這兩個問題都是錯誤的。將FLOOR()添加到datediff結果中，最終得出正確的結果。

最終查詢：

SELECT COUNT(users.UserID), case when users.UserBirthDate IS NULL then 'Sin Registro' 
when FLOOR(datediff(now(), users.UserBirthDate)/365.25) > 50 then '> 51' 
when FLOOR(datediff(now(), users.UserBirthDate)/365.25) > 40 AND FLOOR(datediff(now(), users.UserBirthDate)/365.25) < 51 then '41 - 50' 
when FLOOR(datediff(now(), users.UserBirthDate)/365.25) > 30 AND FLOOR(datediff(now(), users.UserBirthDate)/365.25) < 41 then '31 - 40' 
when FLOOR(datediff(now(), users.UserBirthDate)/365.25) > 26 AND FLOOR(datediff(now(), users.UserBirthDate)/365.25) < 31 then '27 - 30' 
when FLOOR(datediff(now(), users.UserBirthDate)/365.25) > 20 AND FLOOR(datediff(now(), users.UserBirthDate)/365.25) < 27 then '21 - 26' 
when FLOOR(datediff(now(), users.UserBirthDate)/365.25) <= 20 then '< 21' end as AgeGroup FROM users GROUP BY AgeGroup;

正確的結果：

<table><tbody><tr><th>COUNT(users.userid)</th><th>AgeGroup</th></tr><tr><td>4</td><td>21 - 26</td></tr><tr><td>1</td><td>27 - 30</td></tr><tr><td>2</td><td>31 - 40</td></tr><tr><td>4</td><td>&lt; 21</td></tr><tr><td>1</td><td>&gt; 51</td></tr><tr><td>7</td><td>Sin Registro</td></tr></tbody></table>

與在+地板，我得到正確的結果也檢查它。

SELECT COUNT(users.UserID) 
FROM users 
WHERE FLOOR(datediff(now(), users.UserBirthDate)/365.25) > 20 AND FLOOR(datediff(now(), users.UserBirthDate)/365.25) < 27;

返回4，像情況一樣。

我想案件不同於案件的順序。

來源

2016-02-07 16:42:36 RominaV

計算datediff時的差異MySQL

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